CHM 5414 Thermodynamics

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Quiz #2 Solutions

10 October 1996

  1. H2 + F2 arrow right 2 HF is so violent that it can be used to make chemical lasers (from the cascading vibrational energies of the superhot HF). Because it's so exothermic, its equilibrium constant, K, is enormous. Still, the rotational contribution to K, KROT, is modest. Given the molecular parameters, calculate that KROT at 1000°K (where even H2 obeys the integral approximation to zROT). Remember to cancel everything you can or you'll be doing unnecessary work.

    Bond Distances (Å):
    • H2, 0.7417
    • F2, 1.4119
    • HF, 0.9169

zROT = kt/ sigma B
where
B = hbar /2I
and
I = µr² with µ=m1m2/(m1+m2)

KROT = (zHFROT)² / [ zHHROT * zFFROT ]

There will be (kT)² in the numerator and (kT)(kT) in the denominator; they cancel. So it appears as if this KROT is independent of T (since the number of rotational degrees of freedom didn't change).

KROT = ( sigma HH BHH) ( sigma FF BFF) / ( sigma HF BHF)²

= [(2)(2)/(1)²] * (IHF)² / [ (IHH) * (IFF) ]

= 4 [ µHF² / ( µHH * µFF ) ] * [ rHF4 / ( rHH² * rFF² ) ]

= 4 { (19/20)² / [ (1/2) * (19²/38) ] } * (0.9169)4 / [ (0.7414)² * (1.4119)² ]

= 4 (0.19) (0.645)

= 0.49

So it looks like reduced masses have overturned the symmetry factor advantage of HF!

  • In our last quiz, we found the absolute maximum molar entropy of any single degree of freedom to be almost ½ kJ/°K. Estimate the temperature (K) where the He atom approaches its maximum molar entropy. Assume that the pressure can be somehow retained at one atmosphere. What happens long before this T is reached?

    • It's going to be mindlessly enormous, of course. So it won't even be He by the time we've raised it that high; the electrons will have ionized off (now we have 3 free particles) and the nuclei will have decomposed (now we have 6 particles!) and it's possible it's high enough to fracture the nucleons into their component quarks!

    But we have to start somewhere...so we choose the translational entropy of a mole of He atoms. That's actually 3 degrees of freedom, not one; so it'll have about a 1.5 kJ/°K maximal entropy.

    Translational entropy comes from the Sackur-Tetrode equation which problem 19 converts to a constant pressure equation of the form:

    S = R ln (M3/2 T5/2 / P) - 2.316 cal/mol°K
    or
    S = R ln (M3/2 T5/2 / P) - 9.69x10-3 kJ/mol°K

    All we're required to do is solve for T when S, the molar entropy, is 1.5 kJ/mol°K and P=1 (atm).

    Piece of cake. Or rather piece of algebra; expand the logs.

    ln(T) = (2/5) [ ( 1.5 + 0.00969 )kJ/mol°K/R - (3/2)ln(M in grams) ]

    Since R=8.314x10-3 kJ/mol°K,

    ln(T) = 0.4 (181.6 - 2.08) = 71.8 or T = exp(71.8) = 1.5x1031 K (!)

    Yeah, it's no longer helium.

  • The vibrational partition function we've played with is based on the harmonic oscillator model for molecular vibration. What would be the consequences of using a more realistic model (like the Morse function shown in comparison with the H.O. at right) whose stretched potential isn't as rigid as the harmonic oscillator's parabolic curve? For example, at the same temperature and for the same molecule, would zMorse be larger or smaller than zParabolic H.O.? Justify your answer.

    • z measures "availability of states" at a given energy. Imagine an energy near that dissociative tail. The greater width of the Morse function there ensures that its energy levels will be more closely-spaced than those of the H.O. which means there will be MORE available Morse energy levels below any given energy than H.O. levels. That greater density means a larger partition function for Morse.

    5% Bonus Question: What is (CH3)2Hg's sigma ? (D3h)

    Rotations can carry any of the 6 H atoms into any other H atom; so sigma =6.

    We can see that from D3h as E + 2C3 + 3C2' or 1+2+3=6.
    Last modified 15 October 1996